--- title: "Pilot current with tap change (10 MVA)" subject: "Switchgear and Protection" chapter: 7 key: "semester_6/switchgear_and_protection:pilot-current-10mva-tap" source: https://www.bctengineer.com/q/semester_6/switchgear_and_protection/pilot-current-10mva-tap site: BCT Engineering --- # Pilot current with tap change (10 MVA) Switchgear and Protection · Chapter 7: Protective Relays - Asked in: 2071 Magh, 8 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/switchgear_and_protection/pilot-current-10mva-tap - Whole chapter: https://www.bctengineer.com/md/semester_6/switchgear_and_protection/chapter-7 ## Question A 10 MVA, 132/33 kV, star-delta, three-phase, 50 Hz transformer is connected in delta on the low voltage side and in star with star point earthed on the high voltage side. If the CTs on the high voltage have a ratio of 75/1 A, determine the CT ratio on the low voltage side. What would be the current circulating through pilots for a through fault due to which a current of 5 times the full load occurs if the voltage tapping is set to 128 kV at the time of occurrence of fault? ## Short answer LV CT ratio is N = 75 x (132/33)/√3 = 173.2 : 1, star connected. At 128 kV tap and 5 times full load, LV pilot current is 874.77/173.2 = 5.05 A, HV pilot current is √3 x 225.53/75 = 5.21 A, giving 0.158 A relay unbalance. ## Answer HV is star connected, so the HV CTs are connected in **delta**; LV is delta connected, so the LV CTs are connected in **star**. The LV CT ratio is chosen so that pilot currents balance at the nominal ratio 132/33 kV. A tap change upsets this balance and a spill current flows. **Data:** 10 MVA, 132/33 kV; HV CTs 75/1 (in delta); through fault = 5 × full load; tap set at 128 kV. #### Step 1: LV CT ratio For HV current I_H, the delta of HV CTs gives a pilot current √3 I_H / 75. The LV CTs (ratio N:1, star) give I_L / N, with I_L = 4 I_H (132/33). ```text I_L / N = √3 I_H / 75 N = 75 × (I_L/I_H) / √3 = 75 × 4 / √3 = 173.2 LV CT ratio = 173.2 : 1 ``` #### Step 2: Fault currents with 128 kV tap Full-load current on the LV (33 kV) side: ```text I_FL(LV) = 10 × 10⁶ / (√3 × 33 × 10³) = 174.95 A Through-fault LV current = 5 × 174.95 = 874.77 A HV current with tap at 128 kV = 874.77 × 33/128 = 225.53 A ``` #### Step 3: Pilot currents ```text From LV CTs: 874.77 / 173.2 = 5.051 A From HV CTs: √3 × 225.53 / 75 = 5.208 A Out-of-balance (relay) current = 5.208 − 5.051 = 0.158 A ``` **Answer:** LV CT ratio = **173.2 : 1** (star connected). For the through fault with the tap at 128 kV, the pilot currents are about **5.05 A** (LV side) and **5.21 A** (HV side), so an unbalanced current of **0.158 A** circulates through the relay. A percentage-biased relay is needed so that this does not cause tripping. (If "5 times full load" is taken on the HV side, the corresponding values are 5.05 A, 4.90 A and 0.153 A.) ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the LV CT ratio N found? A. Equating LV pilot current I_L/N with the HV delta pilot current sqrt(3) I_H/75 at nominal ratio B. Making N equal to the HV CT ratio C. Making N = 33/132 × 75 D. Dividing 75 by sqrt(3) Answer: A. Pilot currents must balance at nominal ratio, giving N = 75 × 4/sqrt(3) = 173.2. (Key point: Condition for LV CT ratio) 2. The LV CT ratio is: A. 300 : 1 B. 75 : 1 C. 43.3 : 1 D. 173.2 : 1 Answer: D. I_L = 4 I_H, so N = 75 × 4 / sqrt(3) = 173.2. (Key point: LV CT ratio value) 3. Why does a through fault with the tap at 128 kV cause current in the relay? A. The CTs are connected wrongly B. The tap changes the HV current for a given LV current, upsetting the CT balance set for 132/33 C. The fault is internal D. The pilot wires are open Answer: B. CT ratios were matched for 132/33 kV; with 128 kV tap the HV current rises, leaving a spill current. (Key point: Effect of tap change) 4. The out-of-balance pilot current is about 0.158 A. What remedy keeps the relay from tripping? A. Open-circuit the CTs B. Use a percentage-biased differential relay C. Remove the LV CTs D. Increase fault current Answer: B. A biased relay raises its operating level with through current, so the small spill does not trip it. 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